I have the following integral $$g(x_0) = \int_{-\infty}^{\infty} \frac{1}{(1+x^2)^{3/4}}\frac{1}{(1+(x+x_0)^2)^{3/4}}\exp\left(-\frac{2\pi i}{\lambda}\left[\sqrt{1+x^2}-\sqrt{1+(x+x_0)^2} \right] \right)\mathrm{d}x$$

This can be re-written as $$g(x_0) =\int f(x)f^\ast(x+x_0)\mathrm{d}x$$ where $$f(x)=\frac{1}{(1+x^2)^{3/4}}\exp\left(-\frac{2\pi i}{\lambda}\sqrt{1+x^2} \right).$$

Thus, $G(\omega) = |F(\omega)|^2.$

Now, I observe from numerical integration that $g(x_0)$ is very, very, close to $$g(x_0) \approx 2\mathrm{sinc}\left( \frac{2x_0}{\lambda}\right).$$

I would like to understand why. In particular, $F(\omega)$ has a very steep decay (basically brickwall shape - resembling the sinc) around $\omega = \lambda/4$.

If the question, "why is the approximation so good?" too fuzzy, then the reason behind the super-steep decay of $F(\omega)$ can be answered.


  • $\begingroup$ Are you missing a factor? Setting $x_0 = 0$ the integral simplifies to $\int_{-\infty}^\infty \frac{1}{(1 + x^2)^{3/2}} = 2 \neq \mathrm{sinc}(0)$. $\endgroup$ – Willie Wong Aug 24 '16 at 17:01
  • 1
    $\begingroup$ Taking $x_0=1$ and $\lambda=1$ and $5$ I also get disagreement, and not just by a constant factor. Probably Karen copied something wrong. Should we put the question on hold until Karen can correct it? $\endgroup$ – Gerald Edgar Aug 24 '16 at 18:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.